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Problems #7 and #14 on pg.222-223 are graded.
#7, pg.222: Find the coordinate vector $[\vec{x}]_{\mathscr{B}}$ of $\vec{x}$ relative to the given basis $\mathscr{B}=\{\vec{b}_1,\vec{b}_2,\vec{b}_3\}$ where
$$\vec{x}=\begin{bmatrix} 8 \\ -9 \\ 6 \end{bmatrix},$$
and
$$\vec{b}_1=\begin{bmatrix} 1 \\ -1 \\ -3 \end{bmatrix}, \vec{b}_2 = \begin{bmatrix} -3 \\ 4 \\ 9 \end{bmatrix}, \vec{b}_3 = \begin{bmatrix} 2 \\ -2 \\ 4 \end{bmatrix}.$$
Solution: We must find the weights $c_1,c_2,c_3$ that solve
$$c_1 \vec{b}_1 + c_2\vec{b}_2 + c_3\vec{b}_3 = \vec{x},$$
i.e. we must solve
$$c_1 \begin{bmatrix} 1 \\ -1 \\ -3 \end{bmatrix} + c_2 \begin{bmatrix} -3 \\ 4 \\ 9 \end{bmatrix} + c_3 \begin{bmatrix} 2 \\ -2 \\ 4 \end{bmatrix} = \begin{bmatrix} 8 \\ -9 \\ 6 \end{bmatrix}.$$
Set up the augmented matrix and compute its reduced echelon form to get
$$\begin{bmatrix} 1 & -3 & 2 & 8 \\ -1&4&-2&-9 \\ -3&9&4&6 \end{bmatrix} \sim \begin{bmatrix} 1&0&0&-1 \\ 0&1&0&-1 \\ 0&0&1&3 \end{bmatrix}.$$
This yields solution $c_1=-1, c_2=-1, c_3=3$. Thus we have derived the relation
$$(-1) \begin{bmatrix} 1 \\ -1 \\ -3 \end{bmatrix} + (-1) \begin{bmatrix} -3 \\ 4 \\ 9 \end{bmatrix} + 3 \begin{bmatrix} 2 \\ -2 \\ 4 \end{bmatrix} = \begin{bmatrix} 8 \\ -9 \\ 6 \end{bmatrix},$$
which can be observed to be true. Therefore by the defition of coordinate vectors we see
$$[\vec{x}]_{\mathscr{B}} = \begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} -1 \\ -1 \\ 3 \end{bmatrix}.$$
#14,pg.223: The set $\mathscr{B}=\{1-t^2,t-t^2,2-t+t^2\}$ is a basis for $\mathbb{P}_2$. Find the coordinate vector of $\vec{p}(t)=1+3t-6t^2$ relative to $\mathscr{B}$.
Solution: We write $\vec{b}_1=1-t^2, \vec{b}_2=t-t^2,$ and $\vec{b}_3=2-t+t^2$. We must find weights $c_1,c_2,c_3$ such that
$$c_1 \vec{b}_1 + c_2\vec{b}_2 + c_3\vec{b}_3 = \vec{p},$$
or in other words
$$c_1 (1-t^2) + c_2 (t-t^2) + c_3(2-t+t^2) = 1+3t-6t^2.$$
Simplify the left hand side by combining like terms to get
$$(-c_1-c_2+c_3)t^2+(c_2-c_3)t+(c_1+2c_3) = 1+3t-6t^2.$$
Equating coefficients leads to the following system of equations:
$$\left\{\begin{array}{ll}
-c_1-c_2+c_3 &= -6 \\
c_2-c_3 &= 3 \\
c_1+2c_3 &= 1
\end{array}\right..$$
This system can be solved by computing the reduced echelon form of the augmented matrix:
$$\begin{bmatrix} -1 & -1 & 1 & -6 \\ 0&1&-1&3 \\ 1&0&2&1 \end{bmatrix} \sim \begin{bmatrix} 1 &0&0&3 \\ 0&1&0&2 \\ 0&0&1&-1 \end{bmatrix}.$$
Therefore we see
$$[\vec{p}]_{\mathscr{B}} = \begin{bmatrix} 3\\ 2 \\ -1 \end{bmatrix}.$$
